Message-Id: <9503211001.AA27974@st.unocal.com>
From: stssram@st.unocal.com (Bob Myers)
Date: Tue, 21 Mar 1995 02:01:29 -0800
In-Reply-To: Guillemot@aol.com
To: baidarka@imagelan.com
Subject: Re: Paddling...
On Mar 20, 22:30, Guillemot@aol.com wrote:
} Subject: Re: Paddling...
>
> So how about all those vortices? Remember conservation of momentum. Before
> you let go of the mass the momentum was zero. When the mass reaches terminal
> velocity the momentum must still be zero. Momentum (M) = mass * velocity. If
> V= 1 m/sec, the momentum of the mass is 5 kg m/sec. Since the momentum of the
> system must be 0 the momentum of the water must be -5 kg m/sec. So the water
> sits there and spins in little vortices to create enough momentum.
This is not possible. Momentum is a vector. Water spinning in vortices
does nothing to take up linear momentum. Another way of phrasing
"conservation of momentum" is to say, "the center of gravity of the system
is fixed" (well, not accelerating, anyway). If you drop a rock off a cliff,
the earth moves towards the rock just as the rock moves toward the earth.
The momentum transfer occurs only while the weight is accelerating, and
stops when the weight reaches terminal velocity. Since gravity is what
accelerates the weight, the earth itself is part of this system, and that's
where the momentum is transfered to - not the water. Dropping the weight
has very little influence on the water.
All this means is that this invocation of conservation of momentum is
nonsense. Momentum and vortices, as described above, are not related.
> > But you're also right about something else - there *is* a real problem with
> > this force-only analysis. We also need to look at power and energy, and
> > this is where smaller paddles show their inefficiency. In terms of work
> > done (I'm referring to the physics definition of work here), each paddle
> > stroke using the same force pulled through the same arc does the same amount
> > of work. Work = Force x distance. So if you make twice as many paddle
> > strokes, but only result in the same average force on the kayak, you've done
> > twice as much work to do that!
>
> Strokes are not the same as distance. The distance you need to use is how far
> _you_ moved, not how far the paddle moved.
No. Consider the frame of reference of a person sitting in the kayak. He
pulls back on the paddle with one arm, and forward with the other. If he
does this twice as fast with the same stroke and force, he's doing twice as
much work. This is true regardless of whether the boat moves or not.
> > Another way to look at it is the power situation. The power you expend in
> > paddling is equal to the force x velocity -- so if you paddle faster, you're
> > expending more power. Power is also the energy/time, so if your average
> > power expenditure is twice as high, your energy expenditure over that time
> > will also be twice as high.
>
> If you apply the same force to the same mass you will get the same
> acceleration therefore the same velocity.... ad infinitum. The energy is the
> same. Again, you can't look at the velocity of the paddle. You need to look
> at the velocity of the kayak. The paddle is only the conduit through which
> you apply the force. The force from the water is applied to the kayak via the
> paddle. The paddle shaft serves the same function as the rope in the above
> experiment
Yes, you *can* look at the velocity of the paddle. In fact, you have to if
you want to gauge the efficiency of your paddling. The boat is going to
require the same power regardless - I agree. But how you apply that power
makes a big difference for energy efficiency. You have to get the power
from the engine (the person) to the boat/water. If we just look at the
power *required* by the boat, we'll never be able to judge the efficiency of
the power transfer. No energy transfer is 100% efficient.
(Note that it is very important to keep a constant frame of reference - if
we measure the speed of the boat from a stationary position, we have to
measure the speed of the paddle from a stationary position)
You skipped quoting my argument about using the weight. I'll quote it again
here:
> > To get back to your dropping weight thought experiment, if we have to drop
> > the weight more often to get the same time-average force, we have to raise
> > the weight more often - and that takes more energy, without using any more
> > force.
Nick, from your argument about the weight dropping, you agree that the
weight has to drop faster to produce the same force with a pencil than with
a paddle, right? That means the cycle time is shorter for each dropping of
the weight*, and therefore the weight is dropped much more often with a
pencil than with a paddle. Each time the weight is lifted for the next
cycle, the same amount of energy is used to lift the weight. So the total
energy expenditure is equal to the number of times you lift the weight times
the energy to lift it once. Since you're increasing the number of times you
lift it, your energy expenditure has increased.
* Since we're in continuous steady motion. Let's assume it takes no time to
raise the weight again for the next cycle, or even better that there are two
weights, one for each side, and one can be lifted while the other falls, so
we get continuous propulsion.
> > > A wing generates lift perpendicular to it's direction of motion and
> > > perpendicular to it's length. For a paddle to work like a wing it must move
> > > up-and-down and/or in-and-out. Unfortunately when you do that the drag of the
> > > paddle is not parallel to the direction of the kayak and is therefore wasted
> > > energy.
> >
> > Wait a minute - it's not useful to separate lift and drag in this manner.
> > You don't have to move the paddle in either up-and-down or horizontally
> > back; you can combine the two motions. If you do so appropriately, there is
> > no wasted force or energy.
>
> Looking at the vectors. Your paddle is on the left, the blade angle is
> perpendicular to the kayak. You pull it backward towards the right. Drag is
> parallel to motion, so the drag force is forward to the left.
Err - I see what you're talking about now. You're right Nick - I wasn't
thinking of horizontal force perpendicular to the direction of motion. That
component is wasted. I was merely thinking of the vertical components,
which can be compensated for. You can't eliminate the horizontal perpendicular
components, though.
You can make use of lift by up and down motions, though, *without* creating
extra vertical drag and rocking the boat. I don't think it's practical to
do so, though, apart from paddle entry & exit.
Side point about definition of lift:
> Lift is
> produced perpendicular to the blade (not neccessarily perpendicular to drag)
Exactly what definition of lift are you using? Lift is by definition
perpendicular to drag. Lift and drag are merely orthogonal components of
the total hydrodynamic force. Drag is the component in the direction of
motion, and lift is the component perpendicular to the direction of motion.
-- Bob Myers Unocal Tech. & Ops. I. S. Support Internet: Bob.Myers@st.unocal.com P. O. Box 68076 Phone: [714] 693-6951 Anaheim, California 92817-8076